Question

एक आदमी 300 km की दूरी रेल से तय करने मे 2 घंटे कम लेता है यदि रेल की गति सामान्य से 5km/h बडा दी जाए तो सामान्य गति बताऔ

Answer 1

ट्रेन की गति को "x" किमी/घंटा मान लें।

ट्रेन से यात्रा की दूरी = 300 किमी

अब,

समय = दूरी/गति

समय = (300/x) घंटा

बाद में, ट्रेन की गति 5 किमी/घंटा बढ़ जाती है।

अब, नई गति = (x + 5) किमी/घंटा

गति = 300 किमी

समय = 300/(x + 5) घंटा

प्रश्न के अनुसार,

=> (300/x) - 300/(x + 5) = 2

=> 300 [(x + 5 - x)/(x² + 5x)] = 2

=> 300(5)/(x² + 5x) = 2

=> 1500 = 2x² + 10x

=> 2x² + 10x - 1500 = 0

=> x² + 5x - 750 = 0

=> x² + 30x - 25x - 750 = 0

=> x(x + 30) -25(x + 30) = 0

=> (x + 30) (x - 25) = 0

=> x = -30, 25

(-30 खारिज कर दिया क्योंकि गति नकारात्मक नहीं हो सकती)

∴ ट्रेन की गति 25 किमी/घंटा है।

_____________________________

Question :

A man takes less than 2 hours to cover a distance of 300 km by rail. If the speed of the rail is increased by 5 km/h from it's usual(normal) speed. Then find the usual speed of the train.

Solution :

Let speed of train be "x" km/hr.

Distance travelled by train = 300 km

So,

Time = Distance/Speed

Time = (300/x) hr.

Later, the train increase it's speed by 5 km/hr.

Now, new speed = (x + 5) km/hr.

Distance = 300 km

Time = 300/(x + 5) hr.

According to question,

=> (300/x) - 300/(x + 5) = 2

=> 300 [(x + 5 - x)/(x² + 5x)] = 2

=> 300(5)/(x² + 5x) = 2

=> 1500 = 2x² + 10x

=> 2x² + 10x - 1500 = 0

=> x² + 5x - 750 = 0

=> x² + 30x - 25x - 750 = 0

=> x(x + 30) -25(x + 30) = 0

=> (x + 30) (x - 25) = 0

=> x = -30, 25

(-30 rejected as speed can't be negative)

∴ Speed of train is 25 km/hr.

Answer 2

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y= सामान्य गति दें

300/(y + 5) + 2 = 300/y

। । । बईं ओर एक एकल अंश में संयोजित करें

(310 + 2y)/(y + 5) = 300/y

कई गुना पार

310y + 2y ² = 300y + 1500

2y ² + 10y - 1500 = 0

y² + 5x - 750 = 0

(y - 25) (y + 30) = 0

y = -30 ( अमान्य नकारात्मक समय)

y = 25 किमी/घंटा